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User talk:Googology Noob
I suggest you stop with the "SALAD'(X)" function. It won't lead anywhere, as infinite numbers are meaningless on this wiki, by terms of how "big" they are. However, if do you have some actual idea of what you're doing, then go ahead, I'd like to see how this turns out. KthulhuHimself (talk) 15:01, November 8, 2015 (UTC) : I have all the ludicrous salad number components in my head, and I've written them down. My main problem right now is figuring out how to format it on the wiki. Googology Noob (talk) 15:15, November 8, 2015 (UTC) ::Salad numbers, just so you know, are a ''bad thing in googology. Read the wiki article on them. It outright says it's a derogatory term. Don't make large numbers by mashing a bunch of functions together, make your own original system. Cookiefonster (talk) 17:04, November 8, 2015 (UTC) ::::It isn't supposed to be something very serious. I've read quite a lot of the wiki despite not editing, and I am aware of the fact that it is a deregatory term, having read that article quite a few times. ::::It is quite a ludicrous function, as shown by the fact that ''SALAD'(0) is equal to infinity but SALAD'''(-0) is minus infinity (-0?), and ''SALAD'(-1) is the biggest finite number possible (the highlight of "cheating" in googology) + 1. If it's such a big deal I'll remove it from my wiki. Maybe called Googology Noob (talk) 17:13, November 8, 2015 (UTC) Reading my last comment, I see it came out (the way I see it) very passive-aggressive. I'm sorry about that, didn't mean it to come out that way. Maybe called Googology Noob (talk) 17:29, November 8, 2015 (UTC) Already taken Two things: The n-dex numbers already exist, coined under names such as "goggol". n-dex is a suffix that already exists. Look here: http://googology.wikia.com/wiki/-dex KthulhuHimself (talk) 14:28, November 10, 2015 (UTC) :Firstly: Ah, I see. Oh well, I guess I can console myself in that that great minds think alike. I would remove the googolisms that I termed, but I feel that they are useful in explaining. Do you think just stating that those numbers have already been defined (under different names, and linking to the page with that googolism) and they are only for explaining would be OK? :Secondly: Oh dear, I see that n"idex" has been used as well. I'll try and think of something. Thanks for notifying me! The name is Bond. (James "Googology Noob" Bond.) :I think I'll do "-duex". In a way, it sort of fits more due to having "du", a slightly "fuller" suffix, despite being slightly more cumbersome. The name is Bond. (James "Googology Noob" Bond.) :::Sounds good, explanation-sufiixes will help both your progress and your definition-integtiry. :::But for the love of god, change your signature to what it was before. :::KthulhuHimself (talk) 17:51, November 11, 2015 (UTC) :::Okay, thanks! That explanation will be useful in further extensions which I plan to add. By the way, my signature stayed the same, I just manually typed it. Maybe called Googology Noob (talk) 18:25, November 11, 2015 (UTC) :::I actually have another pretty much fully formed (although less extensive) notation in my head and written down, I just need to put it on the wiki. Maybe called Googology Noob (talk) 20:09, November 11, 2015 (UTC) Arrow-recursive Notation I found a way to generalize GoogolX notation: (x,y) = x^y (x,y,z) = x^^^...^^^y with z arrows (x,y,z)->w = [(x,y,(x,y,z))->w-1] (x,y,z)->1 = (x,y,z) (x,y,z)->w->v = [(x,y,z)->(x,y,z)->w->v-1] (x,y,z)->w->1 = (x,y,z)->w These can represent the exponentiation, arrowex, layerex, and stackex googologisms. -- From the googol and beyond -- 18:49, November 13, 2015 (UTC) :Interesting way of putting it, and for the things I've covered so far it is actually a much more comfortable way of expressing it. However, I have some more extensions which will dwarf everything I have previously done and I have thought of another notation quite a while ago which extends the array notation you have shown (and I should really find the time to upload it to the wiki). Thanks for posting it here! Maybe called Googology Noob (talk) 13:04, November 16, 2015 (UTC) Your Notation Interesting Notation. It's not a cascading product of the Up-arrows or Chained-Arrows. It's really creative. On the conterary, you're never going to get to f_{\omega+n}(n) with only using Single arrows /uparrow. You've gotta use degeneration and recursive structures, as shown in BEAF and Chained Arrow Notation. As long as you keep using Arrow notation, I think it's about f_{\omega}(n) in the Fast Growing Hierarchy. As I am the creator of a notation myself (See here), I think again that it has creativity. However, be warned that you must put new structures and functions to make the growth rate stronger. And you might want to change Googolx to Googoly or something because -ex, -dex, -deck, and -ec are commonly used suffixes in Googology. \(\ Antares.H \) 09:10, December 5, 2015 (UTC) :Firstly, thanks for popping by! Secondly, my notation with any amounts of "dimensions", as I called it, already reaches f_w2. 1-D is up arrows, so f_w. 2-D is iterating that, so f_w+1, 3-D is iterating that, so f_w+2, up to N-D, which is f_w+n-1, which has the same growth rate as f_w2. I don't think this notation will ever reach the likes of f_w^w^w or something (and will probably stop much earlier), but that's not horrible. :I like your notation (the concept is nice in my opinion), and good luck with it! I really need to find the time to continue writing the notation, and more important, the willpower to stop playing games and amateurishly programming Turing Machines. 10:43, December 5, 2015 (UTC) :Whoops, forgot to sign in. Maybe called Googology Noob (talk) 14:53, December 5, 2015 (UTC) Question I know that psi_I(n) gives the nth omega fixed point, because "I" diagonalizes over the omega fixed points. Therefore, I is the limit of what we can recursively create with omega fixed points, so it is the limit of the series {psi_I(0), psi_I(psi_I(0)), psi_I(psi_I(psi_I(0)))...}. Is this correct? Maybe called Googology Noob (talk) 06:14, December 20, 2015 (UTC) : This isn't right, for a similar reason as to why \(\Omega\) isn't the limit of \(\psi_\Omega(0),\psi_\Omega(\psi_\Omega(0)),\psi_\Omega(\psi_\Omega(\psi_\Omega(0))),...\). The very point of \(\Omega\) is to be an ordinal so large that you can't reach it using \(\psi_\Omega\) and diagonalization, and similar thing is truw with \(I\) and \(\psi_I\). LittlePeng9 (talk) 09:03, December 20, 2015 (UTC) ::So that means that no matter how many diagonalizations and recursion I apply on \(\psi_I\), I'll never reach I? That's pretty big! Maybe called Googology Noob (talk) 13:04, December 20, 2015 (UTC) Thank you... ...for being the first person on here to actually engage my numbers,though the "BAN" you link to in your comment is a disambiguation page,so I'm not sure which definition you refer to. My illion names,as stated on my first page,started in the 1980s (if prefixes beyond yotta are ever formally adopted I would expand the names to include them,but with "nea","bronto",and "hella" all with their own factions for 1027 I'm not going to take a side at present. The popble function may be unpopular with those who call themselves "googologists",a word I hate as I regard little Milton's naive misnomer for ten sexdecilliards as best disregarded and forgotten in large-number terminology.(Incidentally,I dislike croutons and pick them out of any salad).But I see alloying functions as "standing on the shoulders of giants". Bowers wrote to me that 2 popbled is easily larger than a gongulus while a gongulus popbled is less than a goppatoth. I read a lot of the XKCD Forum "My Number Is Bigger" page and initially devised the Immortal Storm Number as a means of using my illions to generate a truly immense number of nines to use as a power-tower...then I started Moser-gonning those nines and top-loading the tower,and making a succession of numbers that each substituted the previous number for nines.These numbers were arbitrary,but served to develop the Alphabet Number Function stage by stage.I created that function and the Epstein Number Function in 2011 but only got around to publishing the webpage this year. On the numbers that get "nominated",again I admit choices are arbitrary.Every n-factor I create of course increments dimensionally-exponentially slower than the last one. But I'd very much like to see how the Estra Extraordinarily Seriously Nominated Numbers stack up against the various competing hierarchies. --L.E./le@put.com/ 20:49, January 13, 2016 (UTC) :Oh, I was talking about Bird's array notation. Here's my analysis of the first page: :The popble involves quite a few things, but all are incredibly dwarfed by Bower's dimensional arrays, which reach f_w^w^w(n). You repeat that n times, so f_(w^w^w)+1(n). :All your other operators up to pa are dwarfed by it, so that's level f_(w^w^w)+1(n). Re is iterating the iterations, so f_(w^w^w)+3(n). The second Re iterates those repeatedly, so f_(w^w^w)+5. All those extensions of Re reach at maximum f_(w^w^w)+w(n). Not bad, but far from the biggest number ever. :The second page is very arbitrary and hard to read and understand, but it should be about f_(w^w^w)+w2(n). :The third page: :Alphabet number function reaches f_(w^w^w)+w2+2(n) if I'm not mistaken. Your colon bracket function isn't bad: the first part reaches (w^w^w)+w3 (I'm writing only the ordinal now). Epstein's Number Nominating Function is somewhere between (w^w^w)+w4 to (w^w^w)+w5. The end is very hard to analyze, but I believe it is upper bounded by (w^w^w)+w^2, a far cry from the biggest number ever. :The trick to making big numbers is recursion. Now you might say, "That's exactly what I'm doing!", but that's not the right form of recursion. The best way is not to give pages upon pages of explaining how each number recurses, but to give a few rules and let the recursion come by itself. In other words, use recursion to build more recursion. Here's an example from my question mark notation: "An entry consists of 1 or more consecutive question marks in or out of parentheses encased in square brackets. If they are encased in parentheses then there must be at least one question mark before them. A valid expression consists of 2 numbers with one or more entries between them. @@ means 1 or more entries. In the expression a@@b, a is the base and b is the prime. #If there is only one entry and it is a single question mark in square brackets, the expression is equal to base^prime. #If the last entry in the expression is a single question mark encased in square brackets, remove it, nest the expression repeatedly inside the prime (the number of nestings is equal to the prime) and change the prime to the base. #If the last question mark in the last entry is not encased in parentheses, remove it and copy the entry repeatedly after the last current entry. The amount of copies is equal to the prime. #If the last question mark in the last entry is encased in parentheses, and there is more than one question mark inside the parentheses, remove the question mark and copy all of the question marks all of the question marks inside those two parentheses one level of nesting below (outsde those question marks) repeatedly. The amount of copies is equal to the prime. #If the last question mark in the last entry is encased in parentheses, and there is only one question mark inside the parentheses, create repeated consecutive question marks. The amount of question marks is equal to the prime." With these 5 rules I pass far beyond your notation: The expression 9?(?(?))?(?)4 is far bigger than any of your numbers. The reason is that I'm treating each new iteration as a function. As I'm iterating on the latest function, I advance faster. Your notation just iterates a lot on a fixed function, which after barely a few recursions becomes very very weak, and you're "stuck at the same place", so to speak. Maybe called Googology Noob (talk) 13:37, January 14, 2016 (UTC) ::So if I read this wiki's assertions of equivalence correctly,you are claiming that my largest number is somewhere around a Bowers "trilatri".Yet Bowers told me that 2,popbled,exceeded a gongulus...and though he said a gongulus popbled was less than a goppatoth (which is much more than a trilatri) I don't see that a trilatri could be an upper bound to iterations of enormously-repeated,enormously cycled popblings and functions in which popbles are nested.And how is it not recursion to have layers of EEEEnESNNESNN?--L.E./le@put.com/ 17:04, January 14, 2016 (UTC) Let's analyze in depth your notation, and you'll see why it is so. Let's say your notation before the alphabet number function corresponds to f_a(n) in the fast growing hierarchy. Let's compare everything now: A is one iteration on that, so it is f_a(f_a(n). b is two iterations, so it is f_a(f_a(f_a(n))). The mth letter will correspond to f_am+1(n)=f_a(f_a(f_a...f_a(n)...)) with m+1 nestings, so the function g(n) is equal to the nth letter is f_a+1(n)=f_an(n). Did you see that? All that iteration is an increase of 1 in the ordinal. Now, the colon bracket notation is actually a step in the right way. {n:n} will be about f_an(n)=f_a+1(n). However, {n:n:n} will not be f_a+1n(n)=f_a+2(n). You are merely doing 1 iteration of the function, so it will be f_a+12(n). {n::n}={n:n:n:...n:n}=f_a+2(n)=f_a+1n(n). Now we reach a crucial stage. {n:n::n} in your definition is a number with {n::n} Z's. This is a big problem. You're applying a weak function to a big number. You're doing f_a+1(f_a+2(n)), far less than f_a+2(f_a+2(n)), which is in itself far less than f_a+3(n). The trick is not to regard each colon as a fixed function, but as an iterator of the other things. For example, in my notation the expression a??b=a???...?b with b ?'s. I'm not regarding each question mark as a fixed operator, but as something that iterates the rest of the function. In general, it's not hard to see that {nnn}={n:::...::n}=f_a+n(n)=f_a+w(n). Now layers of square brackets iterate that, so it's equal to f_a+w+1(n). Now, you may be asking, why does this equal to only one increase in the ordinal? This is because the "w" (Omega) represents a changing n. It isn't a fixed value, but it is set each time to a new n. That is, f_w+1(3) is not equal to f_3+1(3)=f_4(3), but rather to f_w3(3), that is we iterate f_w(n) 3 times, where each time w is replaced by the n resulting from the previous expression. This is, shortly, very strong. Your first number iterates the f_a+w+1(n) function n times, so it corresponds to f_a+w+2(n). I'll continue the comparison shortly. Maybe called Googology Noob (talk) 18:03, January 14, 2016 (UTC) ::I may be able to refine the bracket-colon notation.In any case,the first Very Extraordinarily Seriously Nominated Number is now defined,I look forward to your further analysis even if not always convinced.--L.E./ 06:31, January 15, 2016 (UTC) I can't see proof that you are bad at Googology. Can you show me proof that you are a googology noob? 21:38, November 14, 2017 (UTC)